[LeetCode] 744. Find Smallest Letter Greater Than Target

Given a list of sorted characters letters containing only lowercase letters, and given a target letter target, find the smallest element in the list that is larger than the given target.

Letters also wrap around. For example, if the target is target = 'z' and letters = ['a', 'b'], the answer is 'a'.

Examples:

Input:
letters = ["c", "f", "j"]
target = "a"
Output: "c"

Input:
letters = ["c", "f", "j"]
target = "c"
Output: "f"

Input:
letters = ["c", "f", "j"]
target = "d"
Output: "f"

Input:
letters = ["c", "f", "j"]
target = "g"
Output: "j"

Input:
letters = ["c", "f", "j"]
target = "j"
Output: "c"

Input:
letters = ["c", "f", "j"]
target = "k"
Output: "c"

Note:

1. letters has a length in range [2, 10000].
2. letters consists of lowercase letters, and contains at least 2 unique letters.
3. target is a lowercase letter.

寻找比目标字母大的最小字母。

给你一个排序后的字符列表 letters ，列表中只包含小写英文字母。另给出一个目标字母 target，请你寻找在这一有序列表里比目标字母大的最小字母。

在比较时，字母是依序循环出现的。举个例子：

如果目标字母 target = 'z' 并且字符列表为 letters = ['a', 'b']，则答案返回 'a'

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/find-smallest-letter-greater-than-target
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

最优解是二分法，当然线性的做法也是可以的。

这里有一个corner case需要判断，就是如果target字母 >= input数组的最后一个字母，那么我们返回的是input数组的首字母。其他环节都是正常的二分法判断。

时间O(logn)

空间O(1)

Java实现

class Solution {
    public char nextGreatestLetter(char[] letters, char target) {
        int len = letters.length;
        // corner case
        if (target >= letters[len - 1]) {
            return letters[0];
        }

        // normal case
        int start = 0;
        int end = len - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (letters[mid] - target > 0) {
                end = mid;
            } else {
                start = mid;
            }
        }
        if (letters[start] > target) {
            return letters[start];
        } else {
            return letters[end];
        }
    }
}

LeetCode 题目总结